\(\begin{array}{l} C = \dfrac{x}{{2x - 2}} + \dfrac{{{x^2} + 1}}{{2 - 2{x^2}}}\\ C = \dfrac{x}{{2\left( {x - 1} \right)}} - \dfrac{{{x^2} + 1}}{{2\left( {{x^2} - 1} \right)}}\\ C = \dfrac{{x\left( {x + 1} \right)}}{{2\left( {x - 1} \right)}} - \dfrac{{{x^2} + 1}}{{2\left( {x - 1} \right)\left( {x + 1} \right)}}\\ C = \dfrac{{{x^2} + x - {x^2} - 1}}{{2\left( {x - 1} \right)\left( {x + 1} \right)}}\\ C = \dfrac{{x - 1}}{{2\left( {x - 1} \right)\left( {x + 1} \right)}}\\ C = \dfrac{1}{{2x + 2}} \end{array}\)
Để $C>0$ thì \(\dfrac{1}{2x+2}>0 \Rightarrow 2x+2>0 \Rightarrow 2x>-2 \Rightarrow x>-1\)
